If omega is a complex cube root of unity,then | vedclass

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(A) We have to calculate $\cos \left(\sum_{k=1}^7(k-\omega)(k-\omega^2) \frac{\pi}{175}\right)$.Since $\omega^2+\omega+1=0$ and $\omega^3=1$,we have $

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$-1$

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(A) We have to calculate $\cos \left(\sum_{k=1}^7(k-\omega)(k-\omega^2) \frac{\pi}{175}ight)$.Since $\omega^2+\omega+1=0$ and $\omega^3=1$,we have $(k-\omega)(k-\omega^2) = k^2 - k(\omega+\omega^2) + \omega^3 = k^2 - k(-1) + 1 = k^2+k+1$.Thus,the expression becomes $\cos \left(\frac{\pi}{175} \sum_{k=1}^7 (k^2+k+1)ight)$.Calculating the sum: $\sum_{k=1}^7 k^2 = \frac{7(8)(15)}{6} = 140$,$\sum_{k=1}^7 k = \frac{7(8)}{2} = 28$,and $\sum_{k=1}^7 1 = 7$.Sum $= 140 + 28 + 7 = 175$.Therefore,$\cos \left(\frac{\pi}{175} 	imes 175ight) = \cos(\pi) = -1$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

If $\omega$ is a complex cube root of unity,then $\cos \left(\sum_{k=1}^7(k-\omega)(k-\omega^2) \frac{\pi}{175}\right) =$

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